Impact of mobile instant messaging applications on signaling load and UE energy consumption

Abstract

Mobile Instant Messaging (MIM) applications transmit not only user-triggered messages (UTMs), but also keep-alive messages (KAMs) via radio access network, which induces heavy burden in control plane channel and wastes user equipment (UE) energy consumption. In this paper, we deduce the joint distribution of KAM period and UTM mean interval from the MIM application traffic characteristics. Correlating the joint distribution with radio resource control (RRC) state machine in LTE networks, we derive two analytical expressions for the control plane signaling load and UE energy consumption respectively. Then, the variation of signaling load and energy usage is demonstrated with different settings of RRC release timer, KAM period and UTM mean interval. The analysis indicates that KAM period is the upper bound of RRC release timer when reducing the signaling load. Besides, five times of UTM mean interval is the upper bound of KAM period when reducing the UE energy consumption and signaling load. These results can guide both network operators and MIM application developers to properly set control parameters for balancing the signaling load and UE energy consumption.

Appendices

Appendix 1: Derivation of the PDF of $t_m$ when $0 \le t_m<t_{{KA}}$

$t_m^n$ is a exponentially distributed random variable. Assume it is a sequence with infinite length of L. The number of times that a specific $t_m^n$ appears in the whole sequence is $L \cdot (\lambda e^{-\lambda t_m^n})$. Therefore, the probability of the specific $t_m^n$ when $t_m^n \ge 0$ can be rewrited as follows:

$$\begin{aligned} f_c(t_m^n)=\lambda e^{-\lambda t_m^n}=\frac{L \cdot (\lambda e^{-\lambda t_m^n})}{L}. \end{aligned}$$

With different value of $t_m^n$, the respective probability can be obtained. Therefore, $f_c(t_m^n)$ can be used to denote the PDF of $t_m^n$. After the transformation of (1), the length of the new sequence, i.e., $L'$ and the number of times that a specific $t_m$ appears in the new sequence, i.e., $n(t_m)$ are no longer L and $L \cdot (\lambda e^{-\lambda t_m^n})$. But both of them can be expressed as the sum of piecewise functions of $f_c(t_m^n)$ when $0\le t_m<t_{{KA}}$:

$$\begin{aligned} \begin{aligned} n(t_m)&=\sum _{j=0}^{\left[ \frac{t_{max}}{t_{{KA}}}\right] }L \cdot f_c(t_m+j \cdot t_{{KA}}),\\ L'&=L+ \sum _{j=1}^{\left[ \frac{t_{max}}{t_{{KA}}}\right] }jL\cdot \left[ \int ^{(j+1)t_{{KA}}}_{jt_{{KA}}} f_c(t_m)dt_m- f_c(jt_{{KA}})\right] , \end{aligned} \end{aligned}$$

where $t_{max}$ is the maximal value of $t_m$. Therefore, the PDF of $t_m$ when $0\le t_m<t_{{KA}}$ can be denoted as follows:

$$\begin{aligned} f(t_m)=\frac{n(t_m)}{l(t_m)}=\frac{\sum _{j=0}^{\left[ \frac{t_{max}}{t_{{KA}}}\right] }L \cdot f_c(t_m+j \cdot t_{{KA}})}{L+\sum _{j=1}^{\left[ \frac{t_{max}}{t_{{KA}}}\right] }j L\cdot \left[ \int ^{(j+1)t_{{KA}}}_{jt_{{KA}}} f_c(t_m)dt_m- f_c(jt_{{KA}})\right] }. \end{aligned}$$

Since $t_{max}$ is much longer than $t_{{KA}}$, we have $t_{max}=N \cdot t_{{KA}}$. Thus, $f(t_m)$ can be simplified as follows:

$$\begin{aligned} \begin{aligned} f(t_m)&=\frac{\sum _{j=0}^{N}f_c(t_m+j \cdot t_{{KA}})}{1+\sum _{j=1}^{N}j\left[ \int ^{(j+1)t_{{KA}}}_{jt_{{KA}}} f_c(t_m)dt_m- f_c(jt_{{KA}})\right] } \\&=\frac{\frac{\lambda \cdot e^{-\lambda \cdot t_m}}{1-e^{-\lambda \cdot t_{{KA}}}}}{\frac{1}{1-e^{-\lambda \cdot t_{{KA}}}}-Ne^{-\lambda \cdot (N+1)\cdot t_{{KA}}}-\left[ \frac{\lambda \cdot e^{-\lambda \cdot t_{{KA}}}}{(1-e^{-\lambda \cdot t_{{KA}}})^2}-\frac{N\lambda \cdot e^{-\lambda \cdot (N+1)\cdot t_{{KA}}}}{1-e^{-\lambda \cdot t_{{KA}}}}\right] }. \end{aligned} \end{aligned}$$

Generally, N is large enough, $f(t_m)$ can be further simplified:

$$\begin{aligned} \begin{aligned} f(t_m) =\frac{\frac{\lambda \cdot e^{-\lambda \cdot t_m}}{1-e^{-\lambda \cdot t_{{KA}}}}}{\frac{1}{1-e^{-\lambda \cdot t_{{KA}}}}-\frac{\lambda \cdot e^{-\lambda \cdot t_{{KA}}}}{(1-e^{-\lambda \cdot t_{{KA}}})^2}} =\frac{\lambda \cdot e^{-\lambda \cdot t_m}}{1-\frac{\lambda \cdot e^{-\lambda \cdot t_{{KA}}}}{1-e^{-\lambda \cdot t_{{KA}}}}}. \end{aligned} \end{aligned}$$

Since $t_{{KA}}$ is much bigger than $1/\lambda$ in most case, $\frac{\lambda \cdot e^{-\lambda \cdot t_{{KA}}}}{1-e^{-\lambda \cdot t_{{KA}}}}$ is negligibly small. Therefore, the final expression of $f(t_m)$ when $0\le t_m<t_{{KA}}$ is:

$$\begin{aligned} f(t_m)=\lambda e^{-\lambda t_m}, \end{aligned}$$

which is the same with $f_c(t_m^n)$.

Appendix 2: Calculation of the mean value of $t_m$ when $0\le t_m\le x~(0\le x<t_{{KA}})$

When $0\le t_m<t_{{KA}}$, the PDF of $t_m$ is typical exponentially distributed. In addition of $0\le x< t_{{KA}}$, the PDF of $t_m$ when $0 \le t_m \le x$ is head-truncated exponentially distributed, which is denoted by $f_t(t_m)$. Since the shape of $f_t(t_m)$ is the same with $f(t_m)$, $f_t(t_m)$ can be expressed as follows:

$$\begin{aligned} f_t(t_m)=\alpha f(t_m), \end{aligned}$$

where $\alpha$ can be calculated by:

$$\begin{aligned} \int ^{x}_{0} \alpha \cdot \lambda e^{-\lambda t_m} dt_m=1, \end{aligned}$$

Therefore, $\alpha$ can be obtained accordingly:

$$\begin{aligned} \alpha =\frac{1}{1-e^{-\lambda x}}, \end{aligned}$$

Then, the mean value of $t_m$ when $0\le t_m\le x~(0\le x<t_{{KA}})$, i.e., h(x) can be calculated as follows:

$$\begin{aligned} h(x) =\int ^{x}_{0}t_m \cdot f_t(t_m)dt_m=\frac{\left( x+\frac{1}{\lambda }\right) \cdot e^{-\lambda x}-\frac{1}{\lambda }}{e^{-\lambda x}-1}. \end{aligned}$$